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Let $X \subset \mathbb{C}P^2$ be a smooth projective curve, and for each $p \in X$, let $T_p$ denote the line tangent to $X$ at $p$.

For any line $L \subset \mathbb{C}P^2$ in the plane, there is a map $X \to L$ that sends $p \in X$ to the unique point of intersection $T_p \cap L \in L$.

How does the topological degree of this map vary with $L$?

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The tangent map $f$ has natural range space $(\mathbb{CP}^2)^*$ the $\mathbb{CP}^2$ of lines in the original $\mathbb{CP}^2$. The degree of the image curve of $f$ is $d(d-1)$ where $d$ is the degree of the original curve. When you ask for the point of interaction with a fixed line $L$, you are considering the mapping from$$(\mathbb{CP}^2)^* \to L$$defined by intersection. Per the discussion, this mapping is not defined at the point $\{L\}$ of $(\mathbb{CP}^2)^*$ corresponding to $L$ itself. However, it becomes a well-defined map$$g: B \to L$$from the blowup $B$ of $(\mathbb{CP}^2)^*$ at the point $\{L\}$. So the degree of the composition $g \circ f$ can indeed change if $\{L\}$ is in the image of $f$. It goes down according to the multiplicity of intersection of the image of $f$ with the $\mathbb{CP}^1$ in $B$ that is collapsed to $\{L\}$ in $(\mathbb{CP}^2)^*$.

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For any line $L \subset \mathbb{C}P^2$ in the plane, there is a map $X \to L$ that sends $p \in X$ to the unique point of intersection $T_p \cap L \in L$.

There is a little problem with your description. If $L$ is itself the tangent line at some $P$ in $X$ then there is no such unique point. For every $Q$ in $L$ there will be some number of tangent lines $T_R(X)$ at points $R$ in $X$ distinct from $P$ through $Q$. There will be a formula if $P$ in $X$ is "fairly general" for this number (something like "class of curve $-$ fixed quantity"), basically trying to fix up how many times we are supposed to count $P$ and $T_P$ in the polar locus. However, this will also go wrong if $P$ in $X$ is more complicated (say a flex point, or a point on a bitangent, or worse).

The problem when $L = T_P(X)$ that I point out above is one way that the degree of the map $X \to L$ can jump down for special $L$, but I certainly don't guarantee that it is the only way.

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    Right, so at least when I was playing with this problem in $\mathbb{R}P^2$, when $L$ is the tangent line to $X$ at $P$, the singularity of the map $X\to L$ at $P$ is removable, by sending $P \mapsto P$. I would expect something similar to work here. My interest is precisely in figuring out if this (and the case when $X \subset \mathbb{C}^2$ is compact and $L$ has only affine intersections with $X$) is the only way the degree can be cut down. – 2017-02-12
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It doesn't (generically). The degree is called the class of the curve; it's the degree of the dual curve.